Method of undetermined coefficients

[yecko]

Homework Statement

y″ − 2y′ − 3y = t ³ e ^5t cos(3t) ---equation 1
Yp = (At³ + Bt² + Ct + D) e ^5t cos(3t) + (Et³ + Ft² + Gt + H) e ^5t sin(3t) ---equation 2

Homework Equations

If there is any term in common, then the entire complex of product that is the choice for Y must be multiplied by t. Repeat as necessary.

The Attempt at a Solution

In equation 2 , "At³e ^5t cos(3t)" is still have same coefficient as the right hand side of equation 1.
why don't we need to times t^4 in order to get rid of it?
thanks

 

[scottdave]

It's been awhile for me but don't you need a form which covers the homogeneous form of solution (the characteristic)?

 

[scottdave]

I did find this. I haven't read through the whole thing yet. https://www.cliffsnotes.com/study-g...tions/the-method-of-undetermined-coefficients

 

[yecko]

I believe I need to find, which the Yh I found is C₁e^3t+C₂e^(-t)...
Just because as this is not mentioned in the notes, so I do not treat it as part of my problem...

Ref: Notes http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf p.14

 

[yecko]

For this from my instructor's notes, exactly explained my problem, but I am totally have no idea what does it about... we always have Ay″+By′+Cy three parts in our questions, but by what principle we have to multiply t^(0/1/2)? why my instructor said 0/1/2 are the only options but not 3 or above?
As for notes by PSU (the link above), "If there is any term in common, then the entire complex of product that is the choice for Y must be multiplied by t. Repeat as necessary." yet this also seems contradict to the example I have given in post 1, for "At³e ^5t cos(3t)" is still have same coefficient as the right hand side of equation 1.
Thanks

 

[Ray Vickson]

View attachment 221810
For this from my instructor's notes, exactly explained my problem, but I am totally have no idea what does it about... we always have Ay″+By′+Cy three parts in our questions, but by what principle we have to multiply t^(0/1/2)? why my instructor said 0/1/2 are the only options but not 3 or above?
As for notes by PSU (the link above), "If there is any term in common, then the entire complex of product that is the choice for Y must be multiplied by t. Repeat as necessary." yet this also seems contradict to the example I have given in post 1, for "At³e ^5t cos(3t)" is still have same coefficient as the right hand side of equation 1.
Thanks

Writing $\cos(w) = (1/2) (e^{iw} + e^{-iw}),$ you can write the right-hand-side of your DE as $t^3 e^{5t} (1/2) (e^{i 3t} + e^{-i 3t})$, so as a sum of two terms of the form $c \, t^3 e^{rt}$, for $r = 5 + 3i$ and $r = 5 - 3i.$. A particular solution $y_p$ of $y'' - 2 y' - 3y = c t^3 e^{rt}$ has form
$$ (A_0 + A_1 t + A_2 t^2 + A_3 t^3) e^{rt} + B e^{3t} + Ce^{-t}.\hspace{2em}(1)$$
Back in the world of "sin" and "cos" you get a sum of two such forms for the two values of $r$. Eventually you get a sum of such forms but with $e^{5t} \cos(3t)$ and $e^{5t} \sin(3t)$ in place of the $e^{rt}.$ In other words, you would get a solution of the form
$$y_p = P_3(t) e^{5t} \cos(3t) + Q_3(t) e^{5t} \sin(3t) + B e^{3t} + Ce^{-t}, \hspace{2em}(2) $$
where $P_3$ and $Q_3$ are two (possibly different) cubic polynomials. It is easy enough to see how starting with (1) you can get (2). However, getting the actual coefficients in $P_3, Q_3$ is lengthy and messy, so be warned.

I did not "guess" the form above; rather, I used a Laplace-transform technique to solve the DE with right-hand-side $t^3 e^{rt}.$ Normally I would solve the entire DE from start to finish using Laplace, but I refrained from doing that just so I could help with your question as it was put to you.

Also: rather than using the "method of undermined coefficients" (which require a certain amount of guesswork and trial-and-error) I would prefer to use the method of "variation of parameters". See, eg.,
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

 

[vela]

Homework Statement

y″ − 2y′ − 3y = t³ e^5t cos(3t) --- equation 1
Yp = (A t³ + B t² + Ct + D) e ^5t cos(3t) + (E t³ + F t² + Gt + H) e^5t sin(3t) --- equation 2

Homework Equations

If there is any term in common, then the entire complex of product that is the choice for Y must be multiplied by t. Repeat as necessary.

The Attempt at a Solution

In equation 2 , "A t³ e^5t cos(3t)" is still have same coefficient as the right hand side of equation 1.
why don't we need to times t^4 in order to get rid of it?
thanks

Suppose the term on the righthand side is $e^{3t}$ (instead of $t^3 e^{5t}\cos 3t$). You might naively guess the particular solution would be of the form $y_p(t) = Ae^{3t}$; however, if you plug this guess into the differential equation, you'd find $y''_p(t) - 2y'_p(t) - 3y_p(t) = 0$. The problem arises because $e^{3t}$ is a solution to the homogeneous differential equation. You modify the particular solution by multiplying by $t$ to get a solution independent of the homogeneous solution.

Do you understand the phrase "in common" now?

 

[yecko]

Writing $\cos(w) = (1/2) (e^{iw} + e^{-iw}),$ you can write the right-hand-side of your DE as $t^3 e^{5t} (1/2) (e^{i 3t} + e^{-i 3t})$, so as a sum of two terms of the form $c \, t^3 e^{rt}$, for $r = 5 + 3i$ and $r = 5 - 3i.$. A particular solution $y_p$ of $y'' - 2 y' - 3y = c t^3 e^{rt}$ has form
$$ (A_0 + A_1 t + A_2 t^2 + A_3 t^3) e^{rt} + B e^{3t} + Ce^{-t}.\hspace{2em}(1)$$
Back in the world of "sin" and "cos" you get a sum of two such forms for the two values of $r$. Eventually you get a sum of such forms but with $e^{5t} \cos(3t)$ and $e^{5t} \sin(3t)$ in place of the $e^{rt}.$ In other words, you would get a solution of the form
$$y_p = P_3(t) e^{5t} \cos(3t) + Q_3(t) e^{5t} \sin(3t) + B e^{3t} + Ce^{-t}, \hspace{2em}(2) $$
where $P_3$ and $Q_3$ are two (possibly different) cubic polynomials. It is easy enough to see how starting with (1) you can get (2). However, getting the actual coefficients in $P_3, Q_3$ is lengthy and messy, so be warned.

I did not "guess" the form above; rather, I used a Laplace-transform technique to solve the DE with right-hand-side $t^3 e^{rt}.$ Normally I would solve the entire DE from start to finish using Laplace, but I refrained from doing that just so I could help with your question as it was put to you.

Also: rather than using the "method of undermined coefficients" (which require a certain amount of guesswork and trial-and-error) I would prefer to use the method of "variation of parameters". See, eg.,
http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

Thank you for your help, yet my homework with this question require to be solved with "method of undetermined equation", so though it is is lengthy and messy, I have to solve it. (I have not learned Laplace transform yet...)
P(t)=Q(t)=( A ₀ + A ₁ t + A ₂ t ² + A ₃ t ³ ) e ^{r t}
and I have got the question of why don't we need to multiply P(t) & Q(t) by t^3?

Suppose the term on the righthand side is $e^{3t}$ (instead of $t^3 e^{5t}\cos 3t$). You might naively guess the particular solution would be of the form $y_p(t) = Ae^{3t}$; however, if you plug this guess into the differential equation, you'd find $y''_p(t) - 2y'_p(t) - 3y_p(t) = 0$. The problem arises because $e^{3t}$ is a solution to the homogeneous differential equation. You modify the particular solution by multiplying by $t$ to get a solution independent of the homogeneous solution.

Do you understand the phrase "in common" now?

Isn't that whenever if the coefficient of the dummy variable is same as the right hand side of the differential equation, it would be canceled out so t/t^2 should be added? how can be determine the situation "in common" without solving it?

Thanks

 

[Ray Vickson]

Thank you for your help, yet my homework with this question require to be solved with "method of undetermined equation", so though it is is lengthy and messy, I have to solve it. (I have not learned Laplace transform yet...)
P(t)=Q(t)=( A ₀ + A ₁ t + A ₂ t ² + A ₃ t ³ ) e ^{r t}
and I have got the question of why don't we need to multiply P(t) & Q(t) by t^3?

Isn't that whenever if the coefficient of the dummy variable is same as the right hand side of the differential equation, it would be canceled out so t/t^2 should be added? how can be determine the situation "in common" without solving it?

Thanks

Do not misrepresent what I wrote. I said that $P(t)$ and $Q(t)$ are possibly-different cubic polynomials, so you get one cubic for $e^{r_1 t}$ and and another cubic for $e^{r_r t}$. When you put things back into trigonometric form, you will have two different cubics: one for $e^{rt} \cos(wt)$ and another for $e^{rt} \sin(wt)$.

You should not multiply $P$ and $Q$ by $t^3$ because that would produce 6th degree polynomials like $A_3 t^6 + A_2 t^5 + A_1 t^4 + A_0 t^3$, and why would you ever want something like that?

About Laplace transforms: it does not matter whether you have seen them; I was just telling you how I arrived at the results, but once you have the results you can verify them by just doing differentiations!

For your final question: if you have $t^n e^{rt}$ on the right, why isn't the solution just something like $t^n e^{rt}$? Well, when you apply the product rule for derivatives, a DE with left-hand-side $a y'' + b y' + cy$ would have terms in which all those derivatives are applied to the $t^n$ part; that would produce terms in $t^{n-1}$ and $t^{n-2}$. You need to cancel these out, and you do that by having something more general that just $t^n e^{rt}$. If you have $P_n(t) e^{rt}$ with an $n$th degree polynomial $P_n$, you can figure out what must be all the coefficients in the polynomial in order that the unwanted terms are canceled out, only to be left with $t^n e^{rt}$ in the end.

 

[yecko]

Do not misrepresent what I wrote. I said that $P(t)$ and $Q(t)$ are possibly-different cubic polynomials, so you get one cubic for $e^{r_1 t}$ and and another cubic for $e^{r_r t}$. When you put things back into trigonometric form, you will have two different cubics: one for $e^{rt} \cos(wt)$ and another for $e^{rt} \sin(wt)$.

You should not multiply $P$ and $Q$ by $t^3$ because that would produce 6th degree polynomials like $A_3 t^6 + A_2 t^5 + A_1 t^4 + A_0 t^3$, and why would you ever want something like that?

About Laplace transforms: it does not matter whether you have seen them; I was just telling you how I arrived at the results, but once you have the results you can verify them by just doing differentiations!

For your final question: if you have $t^n e^{rt}$ on the right, why isn't the solution just something like $t^n e^{rt}$? Well, when you apply the product rule for derivatives, a DE with left-hand-side $a y'' + b y' + cy$ would have terms in which all those derivatives are applied to the $t^n$ part; that would produce terms in $t^{n-1}$ and $t^{n-2}$. You need to cancel these out, and you do that by having something more general that just $t^n e^{rt}$. If you have $P_n(t) e^{rt}$ with an $n$th degree polynomial $P_n$, you can figure out what must be all the coefficients in the polynomial in order that the unwanted terms are canceled out, only to be left with $t^n e^{rt}$ in the end.

let me put another example from my homework,
y''-2y'-15y=e^3xcos(2x)
the solution of the particular integral: y_p=e^3x(Acos(2x)+Bsin(2x))
this won't produce 6th order polynomial if I multiply it with x, which can get rid of being "in common" with the right hand side of the differential integration.
so why won't the solution be multiplied by t?
thanks

 

[Ray Vickson]

let me put another example from my homework,
y''-2y'-15y=e^3xcos(2x)
the solution of the particular integral: y_p=e^3x(Acos(2x)+Bsin(2x))
this won't produce 6th order polynomial if I multiply it with x, which can get rid of being "in common" with the right hand side of the differential integration.
so why won't the solution be multiplied by t?
thanks

Your example is irrelevant to this discussion, and just indicates to me that you do not really understand the material.

If the right-had-side has the form $e^{rx} \cos(wx)$ then of course we would never multiply by $x^k$ (NOT $t^3$ when there is no $t$ in the problem). We multiply by $x^k$ (plus lower-powers) when the right-hand-side has the form $x^k e^{rx}$, etc. Do not make things more difficult for yourself then they need to be.

 

[yecko]

Your example is irrelevant to this discussion, and just indicates to me that you do not really understand the material.

If the right-had-side has the form $e^{rx} \cos(wx)$ then of course we would never multiply by $x^k$ (NOT $t^3$ when there is no $t$ in the problem). We multiply by $x^k$ (plus lower-powers) when the right-hand-side has the form $x^k e^{rx}$, etc. Do not make things more difficult for yourself then they need to be.

is this an exceptional case?
so right hand side with coefficient with x^rx, e^rx, sin(rx), cos(rx),x^rx*sin(rx),x^rx*cos(rx)... all should multiply x/x^2 when in common with particular equation, except "exponential with trigonometric equation" should not multiply x/x^2?
thanks

 

[Ray Vickson]

is this an exceptional case?
so right hand side with coefficient with t^rx, e^rx, sin(rx), cos(rx),t^rx*sin(rx),t^rx*cos(rx)... all should multiply t/t^2 when in common with particular equation, except "exponential with trigonometric equation" should not multiply t/t^2?
thanks

Do you intend $t$ to be the independent variable, or is it $x$? In other words, does your DE have $d/dt$ or does it have $d/dx$? It looks like you are mixing them up and thus confusing not only yourself, but everybody else as well.

 

[yecko]

ok, let me correct it to d/dx:
x^rx, e^rx, sin(rx), cos(rx),x^rx*sin(rx),x^rx*cos(rx) : are they all should multiply x/x^2 when in common with particular equation
and "exponential with trigonometric equation" should not multiply by x/x^2
is that right?

 

[Ray Vickson]

ok, let me correct it to d/dx:
x^rx, e^rx, sin(rx), cos(rx),x^rx*sin(rx),x^rx*cos(rx) : are they all should multiply x/x^2 when in common with particular equation
and "exponential with trigonometric equation" should not multiply by x/x^2
is that right?

Come here with a specific homework problem, and show us your work on it. General questions like the above are not a good fit here.